Question: The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives $16$ years; the standard deviation is $1.7$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a gorilla living between $14.3$ and $19.4$ years.
Solution: $16$ $14.3$ $17.7$ $12.6$ $19.4$ $10.9$ $21.1$ $95\%$ $68\%$ $13.5\%$ $13.5\%$ We know the lifespans are normally distributed with an average lifespan of $16$ years. We know the standard deviation is $1.7$ years, so one standard deviation below the mean is $14.3$ years and one standard deviation above the mean is $17.7$ years. Two standard deviations below the mean is $12.6$ years and two standard deviations above the mean is $19.4$ years. Three standard deviations below the mean is $10.9$ years and three standard deviations above the mean is $21.1$ years. We are interested in the probability of a gorilla living between $14.3$ and $19.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the gorillas will have lifespans within 2 standard deviations of the average lifespan. It also tells us that $68\%$ of the gorillas will have lifespans within 1 standard deviation of the mean. The probability of a particular gorilla living between $14.3$ and $19.4$ years is ${68\%} + \color{orange}{13.5\%}$, or $81.5\%$.